Question

How many bacteria will be present after 5 hours if a culture of bacteria obeys the law of uninhibited growth where if 500 bacteria are present in the culture initially and there are 800 after 1 hour?

Answer 1

THe number of bacteria is `=5243 " bacteria"`

Explanation:

The growth of bacteria is according to the differential equation

`(dN)/dt=kN`

Integration of this equation yields

`ln(N_2/N_1)=k(t_2-t_1)`

or

`(N_2/N_1)=e^(k(t_2-t_1))`

The initial conditions are

`N_1=500` at `t_1=0`

Also,

`N_2=800` at `t_2=1h`

Therefore,

`ln(800/500)=k(1-0)`

Therefore,

`k=ln(8/5)`

Therefore,

After `t_3=5h`, there are

`ln(N_3/N_1)=k(t_3-t_1)`

`ln(N_3/500)=ln(8/5)*5=2.35`

`N_3=500e^(2.35)=5243 " bacteria"`

Answer 2

`color(blue)(5242.88)`

Explanation:

We need to find an equation of the form:

`A(t)=A(0)e^(kt)`

Where:

`bb(A(t))=` the amount after time t.

`bb(A(0)=` the amount at the start. i.e. t = 0.

`bbk=` the growth/decay factor.

`bbe=` Euler's number.

`bbt=` time, in this case hours.

We have:

`A(0)=500`

`A(1)=800`

Using these values:

`800=500e^(k)`

We now solve for `bbk`:

Divide by 500:

`8/5=e^(k)`

Taking natural logarithms of both sides:

`ln(8/5)=kln(e)`

`ln(e)=1` ( The logarithm of the base is always 1 ):

Hence:

`ln(8/5)=k`

And:

`A(t)=500e^(tln(8/5))`

This simplifies to:

`A(t)=500(8/5)^t`

Amount of bacteria after 5 hours:

`A(5)=500(8/5)^5=5242.88`

Answer 3

If it obeys the law of uninhibited growth, growth occurs exponentially .

With `B` as the number of bacteria:

• `B(t) = B_o a^t qquad {(a in RR), (a gt 1):}`

Use the given condition to find `a`:

`B(1) = 800 = 500* a^1`

`implies a = 8/5`

Then after 5 years:

`B(5) = 500 * (8/5)^5 = 5242.88`

Then round that to a whole number of bacteria :)